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3 years ago

What can the student predict if she finds the mass of the whole candy bar, including the wrapper?

Chemistry

1 answer:

Vesna [10]3 years ago

6 0

Answer:

A student can predict Amount of calories present if she finds the mass of whole candy bar,including the wrapper.

<u>Explanation:</u>

Based on the mass of a whole candy bar a student can predict the calories contained in the candy bar. If the mass of candy bar is more than it means it has more calories and if the candy bar has less mass than it shows the presence of fewer calories.

More calories mean more sugar content in the candy bar. With sugar, the candy bar has carbohydrates also. The sugar present in the bar gives us energy.

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3. After 7.9 grams of sodium are dropped into a bathtub full of water, how many grams of hydrogen gas are released?

Pavel [41]

Answer:

3) About 0.35 grams of hydrogen gas.

4) About 65.2 grams of aluminum oxide.

Explanation:

Question 3)

We are given that 7.9 grams of sodium is dropped into a bathtub of water, and we want to determine how many grams of hydrogen gas is released.

Since sodium is higher than hydrogen on the activity series, sodium will replace hydrogen in a single-replacement reaction for sodium oxide. Hence, our equation is:

$\displaystyle \text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}$

To balance it, we can simply add another sodium atom on the left. Hence:

$\displaystyle 2\text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}$

To convert from grams of sodium to grams of hydrogen gas, we can convert from sodium to moles of sodium, use the mole ratios to find moles in hydrogen gas, and then use hydrogen's molar mass to find its amount in grams.

The molar mass of sodium is 22.990 g/mol. Hence:

$\displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}$

From the chemical equation, we can see that two moles of sodium produce one mole of hydrogen gas. Hence:

$\displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}$

And the molar mass of hydrogen gas is 2.016 g/mol. Hence:

$\displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}$

Given the initial value and the above ratios, this yields:

$\displaystyle 7.9\text{ g Na}\cdot \displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}\cdot \displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}$

Cancel like units:

$=\displaystyle 7.9\cdot \displaystyle \frac{1}{22.990}\cdot \displaystyle \frac{1}{2}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1}$

Multiply. Hence:

$=0.3463...\text{ g H$_2$}$

Since we should have two significant values:

$=0.35\text{ g H$_2$}$

So, about 0.35 grams of hydrogen gas will be released.

Question 4)

Excess oxygen gas is added to 34.5 grams of aluminum and produces aluminum oxide. Hence, our chemical equation is:

$\displaystyle \text{O$_2$} + \text{Al} \rightarrow \text{Al$_2$O$_3$}$

To balance this, we can place a three in front of the oxygen, four in front of aluminum, and two in front of aluminum oxide. Hence:

$\displaystyle3\text{O$_2$} + 4\text{Al} \rightarrow 2\text{Al$_2$O$_3$}$

To convert from grams of aluminum to grams of aluminum oxide, we can convert aluminum to moles, use the mole ratios to find the moles of aluminum oxide, and then use its molar mass to determine the amount of grams.

The molar mass of aluminum is 26.982 g/mol. Thus:

$\displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}$

According to the equation, four moles of aluminum produces two moles of aluminum oxide. Hence:

$\displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}$

And the molar mass of aluminum oxide is 101.961 g/mol. Hence: $\displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}$

Using the given value and the above ratios, we acquire:

$\displaystyle 34.5\text{ g Al}\cdot \displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}\cdot \displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}$

Cancel like units:

$\displaystyle= \displaystyle 34.5\cdot \displaystyle \frac{1}{26.982}\cdot \displaystyle \frac{2}{4}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1}$

Multiply:

$\displaystyle = 65.1852... \text{ g Al$_2$O$_3$}$

Since the resulting value should have three significant figures:

$\displaystyle = 65.2 \text{ g Al$_2$O$_3$}$

So, approximately 65.2 grams of aluminum oxide is produced.

5 0

2 years ago

Read 2 more answers

The combustion of 3.795 mg of liquid B, which contains only C, H, and O, with excess oxygen gave 9.708 mg of CO2 and 3.969 mg of

attashe74 [19]

Answer:

Explanation:

9.708 mg of CO₂ will contain 12 x 9.708 / 44 = 2.64 g of C .

3.969 mg of H₂O will contain 2 x 3.969 / 18 = .441 g of H .

mg of O in given liquid B = 3.795 - ( 2.64 + .441 ) = .714 mg

ratio of mg of C , H , O in the compound = 2.64 : .441 : .714

ratio of no of atoms of C , H , O in the compound

= 2.64 / 12 : .441 /1 : .714 / 16

= .22 : .44 : .0446

= .22 / .22 : .44 / .22 : .044 / .22

= 1 : 2 : .2

1 x 5 : 2 x 5 : .2 x 5

= 5 : 10 : 1

empirical formula of the compound = C₅H₁₀O

Volume of 89.8 mL at 1 .00 atm at 200⁰C

volume of gas at 1 atm and 0⁰C = 89.8 x 273 / 473 mL

= 51.83 mL

51.83 mL weighs .205 g

22400 mL will weigh .205 x 22400 / 51.83 g

= 88.6 g

So molecular weight = 88.6

Let molecular formula be (C₅H₁₀O)ₙ

molecular weight = n ( 5 x 12 + 10 + 16 )

= 86 n

86 n = 88.6

n = 1 approx

So molecular formula is same as empirical formula

C₅H₁₀O is molecular formula .

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egoroff_w [7]

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Is a octopus a carnivore or a herbivore

timofeeve [1]

Carnivore, but some are herbivores

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3 years ago

The Bohr model of the atom explained why emission spectra are discrete. It could also be used to explain the photoelectric effec

muminat

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3 years ago

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